Question

How many molecules of N2 are present in 4.086 L of this gas at STP?​

Answer 1

Answer:

2.15 * 10^23 atoms

Explanation:

4.086/22.4 *2* 6.022*10^23

2.043/5.7 * 6.022 * 10^23

2.15 * 10^23 atoms (approx.)

1.07 * 10^23 molecules (approx.)

Answer 2

We know that,

No.of molecules /Avogadro no.=Given volume in litre /22.4 l at stp

Therefore , No. of molecules= 6.022×10^23×4.0486/22.4

=1.09×10^23 molecules.

Hope it helps ....✌

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